Taylor expansion in several variables

A single-variable Taylor series rebuilds a function near a point from its ordinary derivatives: $f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots$ . The moment a function depends on two variables, $f(x,y)$ , the same idea survives almost unchanged — you just replace ordinary derivatives with partial derivatives, and the algebra of "how many ways can I distribute the increment between $x$ and $y$ " turns into a binomial-style pattern involving $f_x$ , $f_y$ , $f_{xx}$ , $f_{xy}$ , $f_{yy}$ , and so on. This matters immediately in MA10001: the second-order Taylor terms about a point are exactly the ingredients of the discriminant test used in the very next topic to classify a critical point as a maximum, minimum, or saddle. Getting comfortable assembling the expansion here pays off directly there.

The core insight

Near $(a,b)$ , $f(x,y)$ is approximated by $f(a,b)$ plus a linear term built from $f_x,f_y$ plus a quadratic term built from $f_{xx},f_{xy},f_{yy}$ , all evaluated at $(a,b)$ — the two-variable analogue of the single-variable Taylor series.

The core idea

Set-up. Let $h = x-a$ and $k = y-b$ be the increments away from the expansion point $(a,b)$ . Assuming $f$ has continuous partial derivatives up to the needed order near $(a,b)$ , the full Taylor expansion is

f(a+h,b+k) = f(a,b) + \left(hf_x + kf_y\right)\Big|_{(a,b)} + \frac{1}{2!}\left(h^2f_{xx} + 2hkf_{xy} + k^2f_{yy}\right)\Big|_{(a,b)} + \cdots,

where: $f_x, f_y$ are first partial derivatives, $f_{xx}, f_{xy}, f_{yy}$ are second partial derivatives (with $f_{xy}=f_{yx}$ by equality of mixed partials for well-behaved $f$ ), and every derivative on the right is evaluated at the fixed base point $(a,b)$ , not at the moving point $(x,y)$ .

The compact operator form. Writing $D = h\dfrac{\partial}{\partial x} + k\dfrac{\partial}{\partial y}$ , the whole series is

f(a+h,b+k) = \sum_{m=0}^{\infty}\frac{1}{m!}\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)^m f\Big|_{(a,b)},

where: the $m$ -th term is expanded formally as a binomial power $(hp+kq)^m$ with $p=\partial_x, q=\partial_y$ , then applied as a differential operator to $f$ . This is why the $m=2$ term picks up the binomial coefficient $2$ on the mixed derivative: $h^2f_{xx} + 2hkf_{xy} + k^2f_{yy}$ , not $h^2f_{xx}+hkf_{xy}+k^2f_{yy}$ .

The truncated form actually used. In MA10001 you almost always stop at second order:

f(a+h,b+k) \approx f(a,b) + hf_x(a,b) + kf_y(a,b) + \frac{1}{2}\left[h^2f_{xx}(a,b) + 2hkf_{xy}(a,b) + k^2f_{yy}(a,b)\right].

Keeping only the first-order terms reproduces the familiar tangent-plane linear approximation from the total differential; adding the second-order terms captures the local curvature of the surface, which is exactly the information the discriminant test needs.

Maclaurin case. When $(a,b)=(0,0)$ , the same formula is called the two-variable Maclaurin expansion, directly paralleling $f(x) \approx f(0)+f'(0)x+\cdots$ in one variable.

Why bother with a two-variable version at all. A function of two variables cannot, in general, be evaluated exactly at every nearby point without repeating the (possibly hard) full calculation — the Taylor expansion lets you do all the hard work once, at a single convenient base point, and then reuse that derivative data to approximate $f$ cheaply at many nearby points, which is exactly what the worked example below does.

Figure

A base point with increments h and k leading to a nearby target point, with a flat tangent plane and a curved surface both drawn to show the linear term versus the linear plus quadratic term

Second-order Taylor expansion for two variables

The goal, given all first and second partial derivatives of $f$ at a base point $(a,b)$ , is to assemble the second-order Taylor approximation $f(a+h,b+k) \approx f(a,b) + hf_x + kf_y + \frac{1}{2}(h^2f_{xx}+2hkf_{xy}+k^2f_{yy})$ correctly, with every term evaluated at $(a,b)$ and the binomial coefficient $2$ correctly placed on the mixed term.

Step 1. Start from the one-variable Taylor idea applied along the straight line $x=a+ht,\ y=b+kt$ for $t\in[0,1]$ , so $t=0$ gives the base point and $t=1$ gives the target point $(a+h,b+k)$ . Define $g(t) = f(a+ht,\ b+kt)$ ; then $g(0)=f(a,b)$ and $g(1)=f(a+h,b+k)$ is exactly the value we want.

Step 2. Expand $g(t)$ as an ordinary single-variable Maclaurin series in $t$ : $g(t) = g(0)+g'(0)t+\frac{1}{2!}g''(0)t^2+\cdots$ , then set $t=1$ so that $f(a+h,b+k)=g(0)+g'(0)+\frac{1}{2!}g''(0)+\cdots$ .

Step 3. Differentiate $g(t)=f(a+ht,b+kt)$ using the chain rule for two variables: $g'(t) = hf_x(a+ht,b+kt) + kf_y(a+ht,b+kt)$ , so $g'(0) = hf_x(a,b)+kf_y(a,b)$ — this produces the first-order term.

Step 4. Differentiate again: $g''(t) = h\big(hf_{xx}+kf_{xy}\big) + k\big(hf_{yx}+kf_{yy}\big) = h^2f_{xx} + 2hkf_{xy} + k^2f_{yy}$ (using $f_{xy}=f_{yx}$ ), all evaluated along the line, so $g''(0) = h^2f_{xx}(a,b)+2hkf_{xy}(a,b)+k^2f_{yy}(a,b)$ — this is exactly where the factor $2$ on the mixed term comes from, since both the $f_{xy}$ and $f_{yx}$ contributions collapse into one term.

Step 5. Substitute $g(0), g'(0), g''(0)$ back into $f(a+h,b+k)=g(0)+g'(0)+\frac{1}{2!}g''(0)+\cdots$ to recover $f(a,b) + hf_x(a,b)+kf_y(a,b) + \frac{1}{2}\big[h^2f_{xx}(a,b)+2hkf_{xy}(a,b)+k^2f_{yy}(a,b)\big]$ , the second-order Taylor expansion.

Worked example

Expand $f(x,y) = x^2y + \sin(x+y)$ about $(1,0)$ up to second order and use it to approximate $f(1.1,-0.1)$ .

Step 1. Compute $f$ and every needed partial derivative, then evaluate at $(a,b)=(1,0)$ : $f(1,0)=\sin 1\approx 0.8415$ ; $f_x=2xy+\cos(x+y)\Rightarrow f_x(1,0)=\cos1\approx0.5403$ ; $f_y=x^2+\cos(x+y)\Rightarrow f_y(1,0)=1+\cos1\approx1.5403$ ; $f_{xx}=2y-\sin(x+y)\Rightarrow f_{xx}(1,0)\approx-0.8415$ ; $f_{xy}=2x-\sin(x+y)\Rightarrow f_{xy}(1,0)\approx1.1585$ ; $f_{yy}=-\sin(x+y)\Rightarrow f_{yy}(1,0)\approx-0.8415$ .

Step 2. Write the increments for the target point $(1.1,-0.1)$ : $h=1.1-1=0.1$ , $k=-0.1-0=-0.1$ .

Step 3. Assemble the second-order approximation:

f(1.1,-0.1) \approx f(1,0) + hf_x(1,0) + kf_y(1,0) + \frac{1}{2}\left[h^2f_{xx}(1,0) + 2hk\,f_{xy}(1,0) + k^2f_{yy}(1,0)\right].

Step 4. Substitute the numbers: $\approx 0.8415 + (0.1)(0.5403) + (-0.1)(1.5403) + \frac{1}{2}\big[(0.01)(-0.8415) + 2(0.1)(-0.1)(1.1585) + (0.01)(-0.8415)\big] = 0.8415 + 0.05403 - 0.15403 + \frac{1}{2}(-0.040) = 0.7415 - 0.020 = 0.7215$ . So $f(1.1,-0.1) \approx 0.7215$ , computed entirely from derivative information at $(1,0)$ , without directly evaluating the (harder-to-handle-by-hand) original expression at the target point.

Competitive stretch

The Maclaurin case is worth memorizing as a shortcut: for $f(x,y)=e^x\sin y$ about $(0,0)$ , the same machinery gives $f(0,0)=0$ , $f_x(0,0)=0$ , $f_y(0,0)=1$ , $f_{xx}(0,0)=0$ , $f_{xy}(0,0)=1$ , $f_{yy}(0,0)=0$ , so the second-order Maclaurin approximation is $e^x\sin y \approx y+xy$ near the origin — matching the product of the known one-variable series $(1+x)(y)=y+xy$ kept to total degree $2$ , a fast sanity check worth running whenever the base point is the origin.

Common traps

Dropping the factor of $2$ on the mixed term. The correct second-order term is $\frac{1}{2}(h^2f_{xx}+2hkf_{xy}+k^2f_{yy})$ ; forgetting the $2$ multiplying $hkf_{xy}$ is the single most common slip.
Evaluating derivatives at the wrong point. Every partial derivative in the formula is evaluated at the fixed base point $(a,b)$ , never at the target point $(x,y)$ — mixing these up produces nonsense answers.
Getting the sign of $h$ or $k$ wrong. If the target point lies "before" the base point in a coordinate, that increment is negative; a sign slip here silently corrupts every later term.
Truncating inconsistently. Once you commit to "second order," every first- and second-order term must appear; dropping one because it looks small changes the accuracy in an uncontrolled way.

Ask the tutor: if assembling the second-order Taylor terms feels error-prone, or you want to see how this connects to the upcoming maxima/minima discriminant test, ask the AI tutor to walk through it step by step.

Why does the mixed second-order term get a factor of 2 but the pure terms h squared f xx and k squared f yy do not?

Because there is only one way to differentiate purely with respect to $x$ twice, or purely with respect to $y$ twice, but there are two equivalent routes to the mixed partial — differentiate with respect to $x$ then $y$ , or $y$ then $x$ — and since $f_{xy}=f_{yx}$ for well-behaved functions, those two equal contributions add up to give the coefficient $2$ . It is exactly the same reason the middle term of $(h+k)^2=h^2+2hk+k^2$ carries a factor of $2$ : the expansion is a binomial power in disguise, applied to derivatives instead of numbers.

Do I need to memorize the full infinite series, or is the second-order truncation enough for this course?

For MA10001 the second-order truncation is what you are expected to use and compute confidently; the full infinite operator series is there mainly so you understand where the pattern comes from and can extend it if a problem explicitly asks for a third-order term. In practice, almost every application here — local approximation, and the discriminant test for maxima and minima that follows this topic — only needs terms up to the quadratic ones, so memorize the second-order formula cold and treat the general series as background understanding.

How is this connected to the maxima and minima test coming up next?

The exact same second partial derivatives $f_{xx}, f_{xy}, f_{yy}$ evaluated at a critical point are what feed the discriminant $D=f_{xx}f_{yy}-f_{xy}^2$ used to classify that point as a local maximum, local minimum, or saddle. The Taylor expansion is the reason that test works: near a critical point the first-order terms vanish, so the local behaviour of $f$ is governed entirely by the sign pattern of the quadratic (second-order) part of its Taylor expansion, which is precisely what the discriminant reads off.